Answer:
The maximum volume of the open box is 24.26 cm³
Step-by-step explanation:
The volume of the box is given as , where and .
Expand the function to obtain:
Differentiate wrt x to obtain:
To find the point where the maximum value occurs, we solve
Discard x=3.54 because it is not within the given domain.
Apply the second derivative test to confirm the maximum critical point.
,
This means the maximum volume occurs at .
Substitute into to get the maximum volume.
The maximum volume of the open box is 24.26 cm³
See attachment for graph.
Idk what unit it is but it’s 80 cubic inches/feet/centimetres
After every drop,the ball bounces to half it's previous height. With that understood.
1st drop -The ball drops 10m
1st bounce - 5m up
2nd drop - 5m down
2nd bounce - 2.5m up
3rd drop - 2.5m down
3rd bounce - 1.25m up
4th drop - 1.25m down
4th bounce - 0.625m up
5th/last drop - 0.625m down
To find the total vertical distance, you add them all.
10+5+5+2.5+2.5+1.25+1.25+0.625+0.625
=29.25m travelled in all.
Answer:
3 students are eating lunches other than salads and sandwiches.
Step-by-step explanation:
To solve this you know that there are 18 students in the cafeteria and 1/6 of them are eating salads and 2/3 are eating sandwiches right? So you would have to think about what 1/6 of 18 is so you know how many students are eating salads, and 1/6 of 18 is 3 so there are 3 students eating salads. Now, you have to find out how many students are eating sandwiches, so you need to know what 2/3 of 18 is. 2/3 of 18 is 12 so now you also know that there are 12 students eating sandwiches. Next, you have to add 12 and 3 and you get 15. Since you know that there are 18 students in the cafeteria, you have to subtract 18 by 15, and you should get 3. So 3 students are eating lunches other than salads or sandwiches.
Hope this helps you! :D